algorithm notes: leetcode#657 robot return to origin

Problem

There is a robot starting at position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.

The move sequence is represented by a string, and the character moves[i] represents its ith move. Valid moves are R (right), L (left), U (up), and D (down). If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.

Note: The way that the robot is “facing” is irrelevant. “R” will always make the robot move to the right once, “L” will always make it move left, etc. Also, assume that the magnitude of the robot’s movement is the same for each move.

Example 1:

Input: "UD"
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.

Example 2:

Input: "LL"
Output: false
Explanation: The robot moves left twice. It ends up two “moves” to the left of the origin. We return false because it is not at the origin at the end of its moves.

Solution

Basic idea

Use a hash table to store the number of ‘R’, ‘L’, ‘U’ and ‘D’ in the string. Traverse the string once and count the number of ‘R’, ‘L’, ‘U’ and ‘D’. Finally, check whether the number of ‘R’ equals to the number of ‘L’ and the number of ‘U’ equals to the number of ‘D’.

Python implemetation

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class :
    def judgeCircle(self, moves):
        """
        :type moves: str
        :rtype: bool
        """
        record = {}
        for move in moves:
            record[move] = record.get(move, 0) + 1
        return record.get('U', 0) == record.get('D', 0) and record.get('L', 0) == record.get('R', 0)

Java implementation

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class  {
    public boolean judgeCircle(String moves) {
        Map<Character, Integer> record = new HashMap<>();
        
        for(char move : moves.toCharArray()){
            int count = record.getOrDefault(move, 0) + 1;
            record.put(move, count);
        }
        
        return (record.getOrDefault('U', 0).equals(record.getOrDefault('D', 0))) &&
               (record.getOrDefault('L', 0).equals(record.getOrDefault('R', 0)));
    }
}

Time complexity analysis

O(N). Becasue we traverse the string once. (N is the number of characters in the string)

Space complexity analsis

O(1).

Link

657. Robot Return to Origin
657. 机器人能否返回原点
(中文版) 算法笔记: 力扣#657 机器人能否返回原点